Problem: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(7t^{4},-5t^{9}\right)$ for time $t\geq 0$. What is the magnitude of the displacement of the particle between time $t=0$ and $t=2$ ? Round to the nearest tenth.
Solution: To find the magnitude of the displacement of the particle, we should first find the particle's horizontal displacement $\Delta x$ and the particle's vertical displacement $\Delta y$. Then we can find the magnitude of the displacement using the distance formula: $\text{Magnitude of displacement }=\sqrt{(\Delta x)^2+(\Delta y)^2}$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between time $t=0$ and $t=2$ : $\Delta x=\int_{0}^{2} 7t^{4}\,dt=\dfrac{224}{5}$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between time $t=0$ and $t=2$ : $\Delta y=\int_{0}^{2} -5t^{9}\,dt=512$ Now we can find the magnitude of the displacement: $\begin{aligned} &\phantom{=}\sqrt{(\Delta x)^2+(\Delta y)^2} \\\\ &=\sqrt{\left(\dfrac{224}{5}\right)^2+\left(512\right)^2} \\\\ &\approx 514 \end{aligned}$ In conclusion, the magnitude of the displacement of the particle between time $t=0$ and $t=2$ is $514$ units.